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combination_with_repetition.cpp
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combination_with_repetition.cpp
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//
// 重複組合せ nHr = n+r-1Cr
//
// cf.
// 高校数学の美しい物語: 重複組合せの公式と例題(玉,整数解の個数)
// https://mathtrain.jp/tyohukuc
//
// よくやる二項係数 (nCk mod. p)、逆元 (a^-1 mod. p) の求め方
// http://drken1215.hatenablog.com/entry/2018/06/08/210000
//
//
// verified:
// ARC 110 D - Factorization
// https://beta.atcoder.jp/contests/abc110/tasks/abc110_d
//
#include <iostream>
#include <vector>
using namespace std;
// 素因数分解
vector<pair<long long, long long> > prime_factorize(long long n) {
vector<pair<long long, long long> > res;
for (long long p = 2; p * p <= n; ++p) {
if (n % p != 0) continue;
int num = 0;
while (n % p == 0) { ++num; n /= p; }
res.push_back(make_pair(p, num));
}
if (n != 1) res.push_back(make_pair(n, 1));
return res;
}
// 二項係数
const int MAX = 210000;
const int MOD = 1000000007;
long long fac[MAX], finv[MAX], inv[MAX];
void COMinit(){
fac[0] = fac[1] = 1;
finv[0] = finv[1] = 1;
inv[1] = 1;
for(int i = 2; i < MAX; i++){
fac[i] = fac[i-1] * i % MOD;
inv[i] = MOD - inv[MOD%i] * (MOD/i) % MOD;
finv[i] = finv[i-1] * inv[i] % MOD;
}
}
long long COM(int n, int k){
if(n < k) return 0;
if (n < 0 || k < 0) return 0;
return fac[n] * (finv[k] * finv[n-k] % MOD) % MOD;
}
//------------------------------//
// Examples
//------------------------------//
// main
int main() {
int N, M;
COMinit();
cin >> N >> M;
auto vec = prime_factorize(M);
long long res = 1;
for (auto pa : vec) {
int k = pa.second;
long long tmp = COM(N+k-1, k); // nHk = n+k-1Ck
res = (res * tmp) % MOD;
}
cout << res << endl;
}